One way to measure ionization energies is ultraviolet photoelectron spectroscopy (PES), a technique based on the photoelectric effect. (Section 6.2) In PES, monochromatic light is directed onto a sample, causing electrons to be emitted. The kinetic energy of the emitted electrons is measured. The difference between the energy of the photons and the kinetic energy of the electrons corresponds to the energy needed to remove the electrons (that is, the ionization energy). Suppose that a PES experiment is performed in which mercury vapor is irradiated with ultraviolet light of wavelength 58.4 nm. (c) The kinetic energy of the emitted electrons is measured to be 1.72 * 10-18 J. What is the first ionization energy of Hg, in kJ>mol?

Video transcript

Hey everyone in this example, it says an electron from zinc metal was removed using 67.3 nanometer beam. Using a photo electron spectroscopy experiment. The emitted electron has a kinetic energy of the following value here and were asked if the difference between the energy of the photons and the kinetic energy of the emitted electron equals the energy involved in the removal of the electron. What is the first ionization energy of our atoms? Inc. So what we want to do is recall that we're going to use our following formula where we take energy set that equal to Planck's constant, multiplied by the speed of light and divided by our wavelength where wavelength or lambda represents the wavelength of our photon. And what we would find is that our energy is going to equal plank's constant, which we would recall is 6.626 times 10 to the negative 34th power units of jewels. Time seconds. We're going to multiply this by the speed of light. Where we recall that the speed of light is equal to a value of three times 10 to the eighth power meters per second. And we're going to divide this by our wavelength of our photon, which according to the prompt, we use a 67. nanometer beam. And we want to recall that we want to convert from nanometers two m since we have meters in our numerator. So we should recall that our prefix nano tells us that we have for one nanometer 10 to the negative ninth power meters. So now we're able to cancel out our units of nanometers we're left with meters. But we can also cancel out meters as well as seconds leaving us with jewels as our final unit of energy here. And this is going to give us a value equal to 2.9536 times 10 to the negative 18th power. And sorry, this should be scooted over jewels per photon. Now, according to the prompt, we have a kinetic energy Of our electron equal to a value of 1.449 times 10 to the negative 18th power jewels. So what we want to do is utilize this value by subtracting it from our energy per photon. So what we would have is 2.953, 6 times 10 To the negative 18th power jewels per photon. And we're subtracting this by the kinetic energy of our electron, which in the prompt is 1.449 times 10 to the negative 18th power jewels per photon. And this difference gives us a value of 1.509 Times 10 to the negative 18th power jewels per photon. So we're going to go ahead and scoot this over. So everything is visible. Now, what we should recognize is that we want our value for energy to be in units of joules per mole or rather for ionization energy. So what we're going to do is take this value that we just found which is our ionization energy and jewels per photon. So we have 1.5 oh nine times 10 to the negative 18th power jewels per photon. And we're going to multiply this to convert from jules two kg jewels by recalling that our prefix kilo gives us 1000 jewels. So we would be able to cancel out our units of jewels. Now now we're at kilo jewels and we want again our unit for ionization energy to be in kilo, joules per moles. So we're going to multiply by another conversion factor where we will go from photons Permal. So we are going to recall avocados number which tells us that we have 6.22 times 10 to the 23rd power atoms or photons in this case per more. This would allow us to cancel our units of photons, leaving us with units of kayla jewels per mole as our final unit for the first ionization energy. And what we're going to get here is a value equal to 906 kayla jewels per mole as our first ionization energy of zinc. And so this would be our final answer. To complete this example. I hope that everything I reviewed was clear if you have any questions leave them down below. Otherwise I'll see everyone in the next practice video